Mutual Inductance is what causes a voltage to appear across the secondary winding of a transformer when an AC voltage source is connected across the transformer's primary winding. A transformer is made up of two wire windings (coils) on either side of a common core. When a current flows through the first or primary winding, it causes a magnetic field to be generated in the core. As the magnetic field fluctuates in strength and polarity with the input AC cycle, it induces a current to be generated in the secondary winding.

For more info see Wikipedia: Transformer

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Reversing a step-down transformer yields a step-up transformer of the same ratio (or more properly, its reciprocal).

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The total resistance of a 10 ohm, a 20 ohm, and a 50 ohm resistor in parallel is 5.9 ohms.

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To calculate the total resistance of \(N\) resistors in PARALLEL, remember that the reciprocal of the \(\text{Total Resistance}\) is equal to the sum of the individual resistors' reciprocal values:

\[\frac{ 1}{ R_{Total}} = \frac{ 1}{R_1} + \frac{ 1}{R_2} + \ldots + \frac{ 1}{R_N}\]

We can re-write that as:

\[R_{Total} = \frac {1}{ \frac{ 1}{R_1} + \frac{ 1}{R_2} + \ldots + \frac{ 1}{R_n}}\]

So for this question:

\begin{align} R_{Total} &= \frac {1}{ \frac{ 1}{10\ \Omega} + \frac{ 1}{20\ \Omega} + \frac{ 1}{50\ \Omega}}\\ &= \frac {1}{ \frac{ 10}{100\ \Omega} + \frac{ 5}{100\ \Omega} + \frac{ 2}{100\ \Omega}}\\ &= \frac {1}{ \frac{ 17}{100\ \Omega}}\\ &= \frac {100}{17}\ \Omega\\ &= 5.9\ \Omega \end{align}

An easy way to check yourself is to remember that total resistance in a parallel network can never be higher than the lowest of the individual resistances. Path of least resistance.

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Another hint: Divide each ohm into the least common denominator (100); then add up the three results and divide the answer into the same denominator.

\begin{align} 100 \div 10 &= 10\\ 100 \div 20 &= 5\\ 100 \div 50 &= 2\\ \\ \text{Total sum} &= 17 \\ 100 \div 17 &= 5.9\\ R &= 5.9 \end{align}

For more info see Wikipedia: Series and parallel circuits

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Remember: Calculating resistance in a series circuit is different than that of a parallel one. The formula for calculating resistance in a parallel circuit is shown below.

\[\frac{1}{R_{Total}}= \frac{1}{R_1} + \frac{1}{R_2}\]

\(\frac{1}{100} + \frac{1}{200} = \frac{3}{200}\)

\(\frac{1}{\frac{3}{200}}=\frac{200}{3}=67 \Omega\)

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The total resistance in a parallel circuit is also equal to the product of all the resistances connected in parallel divided by the sum of all the resistances in parallel, which means you can calculate it with this equation:

\(R=({R_1} \times {R_2})/({R_1} + {R_2})\)

\(R=(100 \times 200)/(100 + 200)\)

\(R=(20000)/(300)\)

\(R≈66.667\)

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To accommodate the higher current of the primary.

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A step up transformer does exactly as the name implies: it "steps up" voltage. The output voltage on the secondary is higher than the input voltage. However, the Conservation of Energy means that the input power on the primary must be higher or equal to the output power on the secondary.

Recall that
\[Power = Current \times Voltage\]

Which we usually write as
\[P = IV \text{ or } P = IE\]

Since the voltage is stepped up, you need extra current to "balance out" the higher voltage on the secondary -- you are trading extra current for the higher voltage.

Higher current means there is more loss over resistive wire. Therefore, to prevent transformer winding wires from overheating, transformer manufacturers make high current winding wire thicker.

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The ratio between primary voltage and secondary voltage is equal to the ratio between primary windings and secondary windings.

The following equation describes a transformer \[\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{I_s}{I_p}\] Where:

• \(N\) = Number of turns
• \(V\) = Voltage
• \(I\) = Current
• \(_p\) = Primary
• \(_s\) = Secondary

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\[\frac{500}{1500} = \frac{120V}{xV}\] \[\frac{120V}{\frac{1}{3}} = 360V\]

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Said another way: 500 turns -> 1500 turns makes it a \(1:3\) step-up transformer, so the output will be 3 times higher. \(120 \times 3 = 360\)

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The impedance ratio of a transformer is equal to the square of its turn ratio.

This means the square root of the impedance ratio of the transformer is equal to its turn ratio.

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1. Find desired impedance ratio \[\frac{600 \Omega}{50 \Omega} = 12\]
2. Take the square root of the impedance ratio to find the turn ratio \[\sqrt{12} \approx 3.464 \approx 3.5\]

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When capacitors are in parallel, you can add the individual capacitances together to get the total equivalent capacitance.

Recall that 1000 picofarads are equivalent to 1 nanofarad.

Converting all the values to the same unit yields:

5.0 nF + 5.0 nF + 0.75 nF = 10.75 nF

For an intuitive explanation of this phenomenon check out Khan Academy's video about capacitors in parallel.

Note that capacitors and resistors behave in opposite ways in this regard: you sum up resistors in series and capacitors in parallel.

Silly hint: "750" is in both the question and the answer. "750" is not used in any of the other distractors.

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The capacitance of three 100 microfarad capacitors connected in series is 33.3 microfarads.

Because each of the capacitors in this problem are of equal value, the total capacitance may be calculated by dividing the common capacitor value (in this case 100 microfarads) by the number, N, of capacitors in the series.

\begin{align} C &= \frac{(Common\ capacitor\ value)}{N}\\ &= \frac{100\ \mu\text{F}}{3}\\ &= 33.3\ \mu\text{F} \end{align}

This is a special case of the generalized equation one can use to calculate the total capacitance of \(N\) capacitors in series:

\[\frac{1}{C_{Total\ Capacitance}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots +\frac{1}{C_N}\]

Which we can also write as:

\[C_{Total\ Capacitance} = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2} + \ldots +\frac{1}{C_N}}\]

One way to think of this is that a capacitor can be two plates separated by a distance, and the bigger the distance, the lower the capacitance. Putting them in series is like making the gap bigger, so the capacitance is lower. Conversely, putting them in parallel is like making the plates bigger, so it increases capacitance.

For more info see Wikipedia: Series and parallel circuits

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(C). The total inductance of three 10 millihenry inductors connected in parallel is 3.3 millihenrys.

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Because the inductors are of equal value, the total inductance (L) of these inductors in parallel may be calculated by dividing the common inductance value (in this case 10 millihenries) by the number of inductors (3).

\begin{align} L &= \frac{\text{Common Value}}{N}\\ &= \frac{10\text{ mH}}{3}\\ &= 3.3\text{ mH} \end{align}

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Alternate Solution:

To solve the general case of this kind of problem, the total inductance of inductors in parallel is the reciprocal of the sum of the individual inductors' reciprocal values as shown:

\[L = \frac{ 1 }{ \frac{1}{L_1} + \frac{1}{L_2} + \ldots + \frac{1}{L_n}}\]

Plugging in the values:

\begin{align} L &= \frac{ 1 }{ \frac{1}{10\text{ mH}} + \frac{1}{10\text{ mH}} + \frac{1}{10\text{ mH}} }\\ &= \frac{ 1 }{ \frac{3}{10\text{ mH}} }\\ &= \frac{10\text{ mH}}{3}\\ &= 3.3\text{ mH} \end{align}

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Silly hint: Just remember II33. Inductance Inductors 3.3. Also, there are 3 letters between I and M for “Millihenries.”

For more info see Wikipedia: Series and parallel circuits

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The total inductance of a 20 millihenry inductor in series with a 50 millihenry inductor is 70 millihenrys.

The total inductance of inductors in series is simply the sum of the individual inductances.

L = L1 + L2 + L3 + ...

So for this question:

L = 20 mH + 50 mH = 70 mH

Note: If all of the inductors were equal in value, the total inductance would be the common inductor value times the number of inductors L = (Common Inductor Value) x N.

Hint: add the values because they're in series.

For more info see Wikipedia: Series and parallel circuits.

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The total capacitance of a 20 microfarad capacitor in series with a 50 microfarad capacitor is 14.3 microfarads.

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The total capacitance of capacitors in series is the reciprocal of the sum of the individual capacitor's reciprocal values as shown:

\[C = \frac{ 1 }{ \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_N}}\]

Solving the problem by this method:

\begin{align} C &= \frac{ 1 }{ \frac{1}{20\ \mu\text{F}} + \frac{1}{50\ \mu\text{F}}}\\ &= \frac{ 1 }{ \frac{5}{100} + \frac{2}{100}}\\ &= \frac{ 1 }{ \frac{7}{100}}\\ &= \frac{ 100 }{ 7 }\\ &= 14.3\ \mu\text{F}\\ \end{align}

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Alternate Method:

When there are only two capacitors in series you can alternately use the simplified equation:

\begin{align} C &= \frac{ C_1 \times C_2 }{ C_1 + C_2 }\\ &= \frac{ 20\ \mu\text{F} \times 50\ \mu\text{F} }{ 20\ \mu\text{F} + 50\ \mu\text{F} }\\ &= \frac{ 1000 }{ 70 }\\ &= 14.3\ \mu\text{F}\\ \end{align}

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Second alternate approach:

Divide each (20 and 50) into the least common multiple (100); add the results together, and then divide that sum into the same number (the LCM).

\begin{align} \frac{100}{20} &= 5\\ \frac{100}{50} &= 2\\ \text{Total sum} &= 7\\ \frac{100}{7} &= 14.3\\ \end{align}

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SILLY HINT: Answer will be lower than the lowest (20) in the series (and .07 is waaaayyy too low!)

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For more info see Wikipedia: Series and parallel circuits

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To increase the total capacitance, you should add a capacitor in parallel.

Capacitors act in an almost opposite manner than inductors and resistors. The total capacitance of capacitors in PARALLEL is the SUM of the individual capacitance values, therefore adding additional capacitors in PARALLEL will result in a larger total capacitance.

For more info see Wikipedia: Series and parallel circuits

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To increase the total inductance, add an additional inductor in series.

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Inductors act in the same manner as resistors in series and parallel circuits. The total inductance in SERIES is the SUM of the individual inductance values. Therefore, adding more inductors in SERIES will increase the total inductance.

For more info, see Wikipedia: Series and parallel circuits.

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